Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $t = \dfrac{-9r + 18}{3r^2 - 21r} \times \dfrac{r^2 - 5r - 14}{-4r + 8} $
First factor the quadratic. $t = \dfrac{-9r + 18}{3r^2 - 21r} \times \dfrac{(r - 7)(r + 2)}{-4r + 8} $ Then factor out any other terms. $t = \dfrac{-9(r - 2)}{3r(r - 7)} \times \dfrac{(r - 7)(r + 2)}{-4(r - 2)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ -9(r - 2) \times (r - 7)(r + 2) } { 3r(r - 7) \times -4(r - 2) } $ $t = \dfrac{ -9(r - 2)(r - 7)(r + 2)}{ -12r(r - 7)(r - 2)} $ Notice that $(r - 2)$ and $(r - 7)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ -9\cancel{(r - 2)}(r - 7)(r + 2)}{ -12r\cancel{(r - 7)}(r - 2)} $ We are dividing by $r - 7$ , so $r - 7 \neq 0$ Therefore, $r \neq 7$ $t = \dfrac{ -9\cancel{(r - 2)}\cancel{(r - 7)}(r + 2)}{ -12r\cancel{(r - 7)}\cancel{(r - 2)}} $ We are dividing by $r - 2$ , so $r - 2 \neq 0$ Therefore, $r \neq 2$ $t = \dfrac{-9(r + 2)}{-12r} $ $t = \dfrac{3(r + 2)}{4r} ; \space r \neq 7 ; \space r \neq 2 $